Chapter 9.6, Problem 49SE

In a hospital’s shipment of 3,500 insulin syringes, 14 were unusable due to defects. (a) At α = .05, is this sufficient evidence to reject future shipments from this supplier if the hospital’s quality standard requires 99.7 percent of the syringes to be acceptable? State the hypotheses and decision rule, (b) May normality of the sample proportion p be assumed? (c) Explain the effects of Type I error and Type II error, (d) Find the p-value.

To determine

State the hypotheses and decision rule.

Answer to Problem 49SE

There is evidence to infer that the proportion of syringes that are effective is 0.997 or more.

There is evidence to infer that the proportion of syringes that are defective is 0.003 or less.

Explanation of Solution

Calculation:

The given information is that, 14 out of 3,500 insulin syringes were unusable due to defects. The number of syringes are effective is 3,486 $\left(=3,500-14\right)$ and the level of significance is $\alpha =0.05$.

Let $\pi$ be the proportion of syringes are effective. That is, $\pi =0.997$.

The test hypotheses are given below:

Null hypothesis:

$H_0:\pi \ge 0.997$

That is, the proportion of syringes that are effective is 0.997 or more.

Alternative hypothesis:

$H_1:\pi <0.997$

That is, the proportion of syringes that are effective is less than 0.997.

The sample proportion is calculated as follows:

$\begin{array}{c}p=\frac{x}{n}\\ =\frac{3,486}{3,500}\\ =0.996\end{array}$

Test statistic:

The formula for test statistic is,

$z_{\text{calc}}=\frac{p-{\pi }_0}{\sqrt{\frac{{\pi }_0\left(1-{\pi }_0\right)}{n}}}$

Where p is sample proportion, $\pi$ is the population proportion and n is the sample size.

Substitute $p=0.996$, ${\pi }_0=0.997$ and $n=3,500$ in the test statistic formula.

$\begin{array}{c}{z}_{\text{calc}}=\frac{0.996-0.997}{\sqrt{\frac{0.997\left(1-0.997\right)}{3,500}}}\\ =\frac{-0.001}{\sqrt{\frac{0.002991}{3,500}}}\\ =\frac{-0.001}{0.0009244}\\ =-1.082\end{array}$

Thus, the test statistic is –1.802.

Critical value:

Software procedure:

Step-by-step software procedure to obtain critical value using EXCEL is as follows:

• Open an EXCEL file.
• In cell A1, enter the formula “=NORM.S.INV(0.05)”
• Output using Excel software is given below:

Decision rule:

If $z_{calc}<-1.645$, then reject the null hypothesis.

Conclusion:

Here, the test statistic is less than the critical value.

That is, $z_{calc}\left(=-1.082\right)>-1.645$.

Therefore, the null hypothesis is not rejected.

Hence, there is evidence to infer that the proportion of syringes that are effective is 0.997 or more.

Let $\pi$ be the proportion of syringes are defective. That is, $\pi =0.003\left(=1-0.997\right)$.

The test hypotheses are given below:

Null hypothesis:

$H_0:\pi \le 0.003$

That is, the proportion of syringes that are defective is 0.003 or less.

Alternative hypothesis:

$H_1:\pi >0.003$

That is, the proportion of syringes that are effective is greater than 0.003.

The sample proportion is calculated as follows:

$\begin{array}{c}p=\frac{x}{n}\\ =\frac{14}{3,500}\\ =0.004\end{array}$

Test statistic:

The formula for test statistic is,

$z_{\text{calc}}=\frac{p-{\pi }_0}{\sqrt{\frac{{\pi }_0\left(1-{\pi }_0\right)}{n}}}$

Where p is sample proportion, $\pi$ is the population proportion and n is the sample size.

Substitute $p=0.004$, ${\pi }_0=0.003$ and $n=3,500$ in the test statistic formula.

$\begin{array}{c}{z}_{\text{calc}}=\frac{0.004-0.003}{\sqrt{\frac{0.003\left(1-0.003\right)}{3,500}}}\\ =\frac{0.001}{\sqrt{\frac{0.002991}{3,500}}}\\ =\frac{0.001}{0.0009244}\\ =1.082\end{array}$

Thus, the test statistic is 1.802.

Critical value:

For right tailed test,

$\begin{array}{c}1-\alpha =1-0.05\\ =0.95\end{array}$

Software procedure:

Step-by-step software procedure to obtain critical value using EXCEL is as follows:

• Open an EXCEL file.
• In cell A1, enter the formula “=NORM.S.INV(0.95)”
• Output using Excel software is given below:

Decision rule:

If $z_{calc}>+1.645$, then reject the null hypothesis.

Conclusion:

Here, the test statistic is less than the critical value.

That is, $z_{calc}\left(=1.082\right)<+1.645$.

Therefore, the null hypothesis is not rejected.

Hence, there is evidence to infer that the proportion of syringes that are defective is 0.003 or less.

To determine

Check whether the normality of the sample proportion may be assumed or not.

Answer to Problem 49SE

The normality of the sample proportion can be assumed.

Explanation of Solution

Calculation:

Rule for normality:

• Rule 1: $n{\pi }_0\ge 10$
• Rule 2: $n\left(1-{\pi }_0\right)\ge 10$

Check the rule:

Rule 1: $n{\pi }_0\ge 10$

$\begin{array}{c}n{\pi }_0=3,500\left(0.997\right)\\ =\text{3,489}\text{.5}\left(>10\right)\end{array}$

Rule 2: $n\left(1-{\pi }_0\right)\ge 10$

$\begin{array}{c}n\left(1-{\pi }_0\right)=3,500\left(1-0.997\right)\\ =3,500\left(0.003\right)\\ =\text{10}\text{.5}\left(>10\right)\end{array}$

Since $n{\pi }_0$ and $n\left(1-{\pi }_0\right)$ is greater than 10, the normality of the sample proportion can be assumed.

To determine

Explain the effects of Type I and Type II errors.

Explanation of Solution

Type I error: Reject the null hypothesis when it is actually true.

A Type I error occurs when the hypothesis test indicates that the proportion of syringes that are effective is less than 0.997. However, in reality, the proportion of syringes that are effective is 0.997 or more. The effect of Type I error is the acceptable shipment would be rejected if the hospital’s quality runs low percentage on insulin syringes.

Type II error: Do not reject the null hypothesis when it is false.

A Type II error occurs when the hypothesis test indicates that the proportion of syringes that are effective is greater than or equal to 0.997. However, in reality, the proportion of syringes that are effective is less than 0.997. The effect of Type II error is the bad shipment would be accepted and due to that, defective syringes are used for insulin injection.

To determine

Find the p-value.

Answer to Problem 49SE

The p-value is 0.1396.

Explanation of Solution

Calculation:

p-value:

$\begin{array}{c}p\text{-value}=\text{NORM}\text{.S}\text{.DIST}\left(z,\text{cumulative}\right)\\ =\text{NORM}\text{.S}\text{.DIST}\left(-1.082,1\right)\end{array}$

Software procedure:

Step-by-step procedure to obtain the p-value using EXCEL is as follows:

• Open an EXCEL file.
• In cell A1, enter the formula “=NORM.S.DIST(–1.082,1)”
• Output using Excel software is given below:

Thus, the p-value is 0.1396.